# Thread: Second order DE help

1. ## Second order DE help

Hi,

I have the equation:
$\begin{cases}y^{\prime\prime} + 3y^{\prime} + 2y = sin~x \\
y(0) = 1 \\
y^{\prime}(0) = 1 \end{cases}$

The auxiliary equation $t^2 + 3t + 2 = 0$ has the solutions
$x_{1}=-1~~ x_{2}=-2$

So the general solution would look like this
$Ae^{-t}+Be^{-2t}$

Now, let $\begin{cases}y = A~Cos(t)+B~Sin(t) \\
y^{\prime} = -A~Sin(t)+B~Cos(t) \\
y^{\prime\prime} = -A~Cos(t)-B~Sin(t) \end{cases}$

gives $-A~Cos(t)-B~Sin(t) + 3\cdot(-A~Sin(t)+B~Cos(t)) + 2\cdot(A~Cos(t)+B~Sin(t))$

Can be simplified down to $3A~Cos(t)-3B~Sin(t) + ASin(t) +B~Cos(t)$

This is where i am stuck, how do you combine this with the general solution?

//Thanks

2. Well, you know that $3A+B=0$ since there are no $cos x$ terms, and $-3B+A=0$ since there is exactly one $sin x$ term.

As a sidebar, it's easier to answer things like this clearly when you use standard notations. Your initial equation indicates $x$ as the independent variable, and $\lambda$ is what is typically used for calculating the characteristic equations.

3. ## That doesn´t look right

If $3A + B = 0$ (1) $-> = B + 3A = 0$
and $B - 3A = 0$ (2) $-> = B - 3A = 0$ then A must be 0

From (1) -> $B = -3A$
Substituting into (2) $-> -3A - 3A = 0 -> -6A = 0$ then A = 0

And from (2) $-> B = 3A$ substituting into $(1) -> B + B = 0$ then B must be 0

Since you have $Ae^-t + Be^-2t$ you would get $y;y`= 0$

4. ## Accually you can

Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???

5. Originally Posted by Henryt999
Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???
I don't know, i don't have the correct answer...

Anyway,

we have $\begin{cases}3A + B = 0 \\
-3B + A = 1\end{cases}$

A = 0.1
B = -0.3

So $y_p = 0.1Sin(t)-0.3Cos(t)$
But how do i pair this up with the general solution?

6. Originally Posted by Jones
I don't know, i don't have the correct answer...

Anyway,

we have $\begin{cases}3A + B = 0 \\
-3B + A = 1\end{cases}$

A = 0.1
B = -0.3

So $y_p = 0.1Sin(t)-0.3Cos(t)$
But how do i pair this up with the general solution?
Just add it to the general solution of the homogeneous equation!

If L(y) is a linear differential operator, then L(y+ y2)= L(y1)+ L(y2). If yh satifies L(yh)= 0 and yp satisfies L(yp)= f(x), then L(yh+ yp)= L(yh)+ L(yp)= 0+ f(x)= f(x). Further, if y is any solution to L(y)= f(x), then L(y- yp)= L(y)- L(yp)= f(x)- f(x)= 0 so that y- yp= yh, a solution to the homogeneous equation. The general solution to the entire equation is the the general solution to the homogeneous equation plus any solution to the entire equation. That's the whole point of this method!